JEE MAIN - Mathematics Hindi (2022 - 28th June Evening Shift - No. 14)

माना   $$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+2 \hat{j}-\hat{k}$$   तथा   $$\vec{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}$$ हैं, जहाँ $$\alpha \in \mathbf{R}$$ है। यदि समांतर चतुर्भुज, जिसकी संलग्न भुजाएँ   $$\overrightarrow{\mathrm{a}}$$   तथा   $$\overrightarrow{\mathrm{b}}$$   हैं, का क्षेत्रफल $$\sqrt{15\left(\alpha^{2}+4\right)}$$ है, तो $$2|\overrightarrow{\mathrm{a}}|^{2}+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})|\overrightarrow{\mathrm{b}}|^{2}$$ का मान बराबर है :
10
7
9
14

Explanation

$$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ तथा $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$

$$\therefore$$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 2 & { - 1} \cr { - 2} & \alpha & 1 \cr } } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$$

अब $$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)} $$

$$ \Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$$

$$ \Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$$

$$\therefore$$ $$\alpha = \, \pm \,3$$

अब, $$2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$$

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